Difference between revisions of "2019 AMC 10A Problems/Problem 7"
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==Solution 2== | ==Solution 2== | ||
− | Like in Solution 1, let's first | + | Like in Solution 1, let's first work out the slope-intercept form of all three lines: |
− | <math>(x,y)=(2,2)</math> and <math>y=x | + | <math>(x,y)=(2,2)</math> and <math>y=\frac{x}{2} + b</math> implies <math>2=\frac{2}{2} +b=1+b</math> so <math>b=1</math>, while <math>y=2x + c</math> implies <math>2= 2 \cdot 2+c=4+c</math> so <math>c=-2</math>. Also, <math>x+y=10</math> implies <math>y=-x+10</math>. Thus the lines are <math>y=\frac{x}{2} +1, y=2x-2,</math> and <math>y=-x+10</math>. |
− | Now we find the | + | Now we find the intersection points between each of the lines with <math>y=-x+10</math>, which are <math>(6,4)</math> and <math>(4,6)</math>. Applying the [[Shoelace Theorem]], we can find that the solution is <math>6 \implies \boxed{\textbf{(C) }6}.</math> |
==Solution 3== | ==Solution 3== | ||
Line 19: | Line 19: | ||
==Solution 4== | ==Solution 4== | ||
− | We can graph the points or use the methods above to find the coordinates, which are <math>(4, 6)</math>, <math>(6, 4)</math>, and <math>(2,2)</math>. Using the [[Shoelace Theorem]], we find that it is equal to <math>(4\cdot 2)-(2\cdot 6)+(2\cdot 4)-(6\cdot 2)+(6 \cdot 6)-(4\cdot 4) = 12</math>. Because the Shoelace Theorem tells us to find the half of that sum, we get <math>\boxed{\ | + | We can graph the points or use the methods above to find the coordinates, which are <math>(4, 6)</math>, <math>(6, 4)</math>, and <math>(2,2)</math>. Using the [[Shoelace Theorem]], we find that it is equal to <math>(4\cdot 2)-(2\cdot 6)+(2\cdot 4)-(6\cdot 2)+(6 \cdot 6)-(4\cdot 4) = 12</math>. Because the Shoelace Theorem tells us to find the half of that sum, we get <math>\boxed{\textbf{(C) }6}</math>. |
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==Solution 5 (Draw it out)== | ==Solution 5 (Draw it out)== | ||
− | We draw the | + | We draw the three lines given in the question on grid paper. It can be easily seen that the three points of intersection are <math>(2, 2)</math>, <math>(4, 6)</math> and <math>(6, 4)</math>. After drawing a square with vertices <math>(2, 2)</math>, <math>(2, 6)</math>, <math>(6, 6)</math> and <math>(6, 2)</math>, we get that the triangle's area is equal to <math>16-4-2-4=6</math>. Thus the answer is <math>\boxed{\textbf{(C) }6}</math>. |
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==See Also== | ==See Also== |
Revision as of 21:06, 17 February 2019
- The following problem is from both the 2019 AMC 10A #7 and 2019 AMC 12A #5, so both problems redirect to this page.
Contents
Problem
Two lines with slopes and
intersect at
. What is the area of the triangle enclosed by these two lines and the line
Solution 1
The two lines are and
, which intersect the third line at
and
. So we have an isosceles triangle with base
and height
.
Solution 2
Like in Solution 1, let's first work out the slope-intercept form of all three lines:
and
implies
so
, while
implies
so
. Also,
implies
. Thus the lines are
and
.
Now we find the intersection points between each of the lines with
, which are
and
. Applying the Shoelace Theorem, we can find that the solution is
Solution 3
Like the other solutions, solve the systems to see that the triangles two other points are at and
. The apply Heron's Formula. The semi-perimeter will be
. The area then reduces nicely to a difference of squares, making it
Solution 4
We can graph the points or use the methods above to find the coordinates, which are ,
, and
. Using the Shoelace Theorem, we find that it is equal to
. Because the Shoelace Theorem tells us to find the half of that sum, we get
.
Solution 5 (Draw it out)
We draw the three lines given in the question on grid paper. It can be easily seen that the three points of intersection are ,
and
. After drawing a square with vertices
,
,
and
, we get that the triangle's area is equal to
. Thus the answer is
.
See Also
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 4 |
Followed by Problem 6 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.